

A156769


a(n) = denom(2^(2*n2)/factorial(2*n1)).


13



1, 3, 15, 315, 2835, 155925, 6081075, 638512875, 10854718875, 1856156927625, 194896477400625, 49308808782358125, 3698160658676859375, 1298054391195577640625, 263505041412702261046875
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OFFSET

1,2


COMMENTS

Resembles A036279, the denominators in the Taylor series for tan(x). The first difference occurs at a(12).
The numerators of the two formulas for this sequence lead to A001316, Gould's sequence.
Stephen Crowley indicated on Aug 25 2008 that a(n) = denom(Zeta(2*n)/Zeta(12*n)) and here numer((Zeta(2*n)/Zeta(12*n))/(2*(1)^(n)*(Pi)^(2*n))) leads to Gould's sequence.
This sequence appears in the Eta and Zeta triangles A160464 and A160474. Its resemblance with the sequence of the denominators of the Taylor series for tan(x) led to the conjecture A156769(n) = A036279(n)*A089170(n1).  Johannes W. Meijer, May 24 2009


LINKS

Table of n, a(n) for n=1..15.


FORMULA

a(n) = denom(product(2/(k*(2*k+1)), k=1..n1)).
G.f.: (1/2)*z^(1/2)*sinh(2*z^(1/2)).


MAPLE

a := n >(2*n1)!*2^(add(i, i=convert(n1, base, 2))2*n+2); # Peter Luschny, May 02 2009


CROSSREFS

Cf. A036279 Denominators in Taylor series for tan(x).
Cf. A001316 Gould's sequence appears in the numerators.
From Johannes W. Meijer, May 24 2009: (Start)
A160464 and A160474 are the Eta and Zeta triangles.
Equals abs(A117972(n))/A000265(n)
Equals A036279(n)*A089170(n1)
Cf. A160469 (which resembles the numerators of the Taylor series for tan(x)).
(End)
Sequence in context: A090627 A070234 A036279 * A029758 A103031 A012474
Adjacent sequences: A156766 A156767 A156768 * A156770 A156771 A156772


KEYWORD

easy,nonn


AUTHOR

Johannes W. Meijer, Feb 15 2009


STATUS

approved



